 So let's start trying to solve quadratic equations, and we'll start by doing this the hardest way possible. And that is to say what we'll do is we'll try to factor them. Now factoring is actually based on something that's rather important, known as the zero product property of the real numbers. And that comes down to the following. If I have a product equal to zero, then what I know for certain is that one of the two factors is equal to zero. So that if I have a product equal to zero, I know that either A is equal to zero or B is equal to zero. Now this means that if I can factor a quadratic expression equal to zero, a quadratic equation, I can reduce that equation into two simpler equations. One factor is zero. The other factor is zero. The thing to keep in mind is that factoring is the hardest and least efficient way to solve quadratic equations. Outside of very, very, very, very, very, very carefully constructed problems, given in a math class, specifically devoted to solving quadratic equations by factoring, you will almost never be able to solve quadratic equations this way. So you might wonder why we spent so much time talking about factoring and solving quadratic equation this way. Well, the answer to that question is, well, actually I don't have a good answer to that question. There isn't any really good reason why we'd want to try to solve quadratic equations by factoring. But we do anyway because we do make these very, very, very, very carefully constructed problems. So let's take an example. Say I want to solve the quadratic equation, x squared minus 6x equal to 7, and an important thing to notice is that in order for this to work, we have to have product equal to zero. So even if I factor the left-hand side, it's not equal to zero, so I can't use the zero product property. And again, the other thing to keep in mind is most quadratic expressions can't be factored. So in general, it is a waste of time even thinking about factoring, unless you know in advance that the question can be answered this way. So factoring does require that we have product equal to zero, so I need to make sure that whatever I have, whenever I do the factorization, I'm going to get zero. So I need to make sure that I have equals zero, not equal seven. So I'll get all the terms over to the left-hand side by subtraction. And so now, a little analysis goes a long way and certainly saves us from wasting a lot of effort. If I should factor the left-hand side, that product will be equal to zero. So now I have a useful form. It's at least insofar as factoring itself is worthwhile. It's worthwhile to consider factoring. So let's think about that. In order to produce a quadratic by multiplying two linear expressions, I'm going to multiply something like ax plus b, times cx plus d. And when I expand that out, I get acx squared plus bcx plus adx plus bd. And what that says is that if I want to obtain this expression by multiplying two things like this together, then my product ac has to give me one, my coefficient of x squared. My product bd has to give me my constant term negative seven. So I need to find two things ac that multiply to one, and two things bd that multiply out to negative seven. So I can find two numbers that multiply to one. How about a equals, well, no sense of being too complicated about this. Let's try a equals one and c equals one. So there's my ax one x, cx one x. And I need two things bd that multiply out to negative seven. So again, I can try b equals seven, d equals negative one, and I can find those products. So b equals seven, d equals negative one, and I'll drop those in. And so I have x plus seven times x minus one equals zero. And because our first random guess at an answer is always the correct one, we can immediately go on to the next step. Well, actually, our first random guess might not actually be what we need. So we do need to verify that this expression here does, in fact, factor as x plus seven times x minus one. So how can we do that? Well, if that's true, then the product of these two should give us our x squared minus x minus seven. So I'll multiply those out. And I find I get product x squared plus six x minus seven, which is not the expression that I wanted. I wanted x squared minus six x minus seven. I got x squared plus six x minus seven. So if I take one step beyond this line here, I'm solving a different problem, and I'm going to get the wrong answer. So what do I have to do? Well, let's try again. So again, I want to find two numbers that multiply to one. So I could try again a equals one, c equals one. And I also want two numbers that multiply to negative seven. And this time I'll try negative seven plus one. And so I'll get this as a potential factorization. I'll again check to make sure that I have the same expression. So I multiply those out. After all the dust settles, I get x squared minus six x minus seven. That is what I'm looking for. And so this is the correct factorization. And now I have product equal to zero. And so now I can solve it. So I know that one of the two factors is zero. So either x minus seven is, and that gives me solution x equals seven, or x plus one is equal to zero. And that gives me solution x equals negative one. Now it's important to recognize that both of these are actually a solution. Both factors will give us a solution. So I should say that my solutions are x equals seven or x equals negative one. Point of grammar here. We do want to use this term or we don't want to say that the solutions are x equals seven and x equals negative one. Because that's implying that x is both seven and at the same time it's negative one. You do want to specify or in our answer. So let's take another problem. Solve by factoring 6x squared plus 11x minus 10 equal to zero. So here we have something equal to zero. If I factor this I'll have product equal zero and that will allow me to resolve the equation. So let's see. So again we'll make the assumption that this factors as something of the form ax plus b times cx plus d. And that says that ac has to give me my six, my coefficient of x squared. Bd has to give me my constant negative 10. So I need to find two numbers a and c that multiply to six and two numbers b and d that multiply to negative 10. So well for example one six that multiplies to six, one negative 10 that multiplies to negative 10. Since our first random guess at an answer is always correct we can use this. No we should actually multiply this out and make sure that it works. So I'll multiply that out and I get 6x squared minus 4x minus 10 and that is not what I want. So this factorization is useless. I could try a different. So I could switch those numbers around minus 1 and 10. Again I multiply this out and I find that I get not what I want. I get something else. And so again this factorization is useless. And so let's see. Well how about two other things that multiply to negative 10? How about two and negative five? Here is a potential factorization. I multiply them out and I get not what I want. Again useless. And I'll switch things around and I get not what I want. And I'll try something else. So again I need two things that multiply to six. So maybe I'll try two and three negative one and 10. So here my two numbers multiply to six. My two numbers multiply to negative 10. I multiply them out and I get not what I want. So again a useless factorization. And I keep trying and I keep trying and I keep trying. And well this one's close. This one's close. This is almost what I want. Now the thing you might notice here is that if I switch where those plus and minus are, if I keep the coefficients the same but I switch the plus and minus, that changes the sign of the middle term. So here I'm close. But I want a plus 11x and not a minus 11x. So this is close. And it suggests that if I switch the location of the plus and minus, I am going to get what I need. And finally, after all of this, I get the factorization. And so when you make that decision to solve by factoring, you condemn yourself to going through all of this until you find the factors. There is no alternative. There's no alternative to going through that entire list until you find the factor. But assuming that you have all the time in the world and all the patience that you could possibly need for this, we find the factorization and there's our factorization. So let's see. Well, I have product equal to zero. So now I can solve this by setting each factor individually equal to zero. 3x minus 2 is zero. So I'll solve that. x equals 2 third is one solution. The other equation, 2x plus 5 equals zero. I'll solve that. x equals minus 5 halves. And that gives us our two solutions. And again, solving by factoring means that you have to have all the time and patience required in order to get this factorization here. We had to go through all of this. And there is no shortcut. So the question you might want to know is, well, do I really have to solve quadratic equations by factoring? Isn't there an easier way than going through this horrible mess? Isn't there an easier way to solve quadratic equations? And the answer to that is absolutely. And we'll take a look at an easier method of solving quadratic equations next.